breaking a large square into smaller parts the best way
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The famous squaring the square problem, solved by Tutte (and others) reduces squaring the square to an electrical network problem about finding a certain network of resistors.
en.wikipedia.org/wiki/Squaring_the_square
🟩 Simple proof: No matter the layout, every horizontal line passes through a linear combination of 40s and 49s. If every line is the same (linear combination) then use vertical lines instead. Consider any two differing lines. These are two /different/ linear combinations of 40 and 49 that add up to the same number. The smallest such number is 40x49 = 1960, composed of all 40s, or all 49s. Choosing all 49s plus one row and one column of 40s makes 2000.
do there have to be two different linear combinations? take n=2×40×49 and you can build the square out of 4 blocks of 40×49 sized squares, build out of just one type. Then every row and every column has 49 40-squares and 40 49-squares.
That's not the smallest solution, but it is one that contradicts your assumption ;-)
@@deinauge7894 🤔 Exercise for the reader. ;)
Hey Michael, great video as always
I just wanted to make a comment of something not so obvious to me, when you say n = 40a + 49b, you exclude the case a=0 or b=0, and because of that you can find the smallest cases by substituting a=1 b=40 or a=49 b=1, or else the smallest case would have been a=1 b=0.
I know that we have to use one of each square, and therefore the case I found is impossible, but I initially I didn't see why it couldn't have been b=0 and a bigger.
Finally I understood, necessarily n can be expressed as 40a+49b with a and b both bigger than 0, because if you counted along the side of the square and found only squares of side 40 on all sides (hence b=0) that would mean you could form a smaller square by removing the outer layer of 40x40 squares from 2 sides. Same reasoning works for showing a>0.
In the final solution, you can count along the side of the square and find only squares of side 40 on all sides. Your argument breaks down because removing the outer layer of 40x40 squares from 2 sides might not leave behind any 40x40 squares.
Instead, you can note that there must exist either a row or a column that contains both a 40x40 square and a 49x49 square. (If not, then we cannot have both kinds of square.)
@@zygoloid if you count along *all* sides and find only 40 side squares, then removing them from only 2 sides will not eliminate all of them since there are 2 more sides we did not remove.
Oh, sure. (Prior to your comment edit it looked like you were suggesting counting along only two of the sides.)
It seems that many others weren't entirely sure about this solution, but I found this video to be one of my favorites in quite a while! The use of roots of unity comes completely out of left field (i.e., is unexpected, to our non-American idiom friends), but it makes the argument so elegant! 👍👍
Sounds like me in the supermarket buying cheese. For some reason, in the UK, cheese commonly comes in blocks of 350 g or 550 g, so I am calculating which is the better buy, 7 of the 550 g or 11 of the 350 g.
Why do you need 7 of 550 g of cheese mate?
@tommasoantonelli7176 A lot of fondeaus to make. No, I'm just trying to figure out which is the better value so working to the lowest common multiple.
The price tag on the shelf should have the price per unit weight (e.g. £/kg) on it which will easily let you compare values.
@@reubenmckayWhere is the fun in that.
@@tommasoantonelli7176 -- There is no such thing as "cheese mate." If you are going to address "mate," then you need to put a comma right before "mate."
"Why do you need ... of cheese, mate?"
We have that lcm(40, 49) = 1960. But this can only be filled with squares of a single size. To have both sizes used, it needs to be padded on two adjacent sides by squares not yet used. (That's why we need to use the lcm, so we can evenly line the 1960×1960 square with the other size.) Those can be 40×40 (resulting in n=2000) or 49×49 (resulting in n=2009).
The whole excursion through the roots of unity seems kind of a roundabout way of getting there.
There's a few assertions you made there that are sort of obvious but unproven. The point of the roots of unity trick is to prove them.
Using complex numbers baked in symmetry.
There needs to be an argument that no assymetric tiling will be smaller.
16:29
Hi,
Like many of us, I did not understand the connection with the nth roots of the unity.
Besides the fact that 1+b+b^2+b^3+ ... + b^39 is not zero, what is zero is up to b^48 .
I'm way more bothered by 4Ях49 than I should be
The elegance of the backwards R symbol here marred by the pointed 4
This is awesome. Wow.
I guess it's pretty obvious that at least one of 4 sides of the composite square should consist of mono 40 or 49 squares (otherwise it's not smallest). Having this we can solve it fast
it’s “pretty obvious” but can you prove it?
Without knowing the roots-of-unity trick, what could motivate that initial step?
This is a generalisation of a simpler trick:
Imagine a chessboard with two opposite corners removed. Can it be covered with 2x1 dominoes?
The answer is no: each domino covers one white square and one black square, and there are not equal numbers of white and black squares.
To make the colouring argument a bit more manageable for larger numbers of colours, we can treat black as 1 and white as -1 (square roots of unity) and observe that a board is only tileable if its sum is zero. Then we can generalize to Mx1 tiles by using Mth roots of unity instead.
With that background, I think this approach to this problem is a bit more natural: the numbering used here corresponds to a tiling with a combination of 1x40 and 49x1 tiles.
I understood each small step, but I have no idea why this worked as an approach when zooming out looking at the bigger picture :D Good video nonetheless!
One question Michael, when you conclude that n is of the form 40a+49b, why did you exclude the case a=b=1? It was my first and simplest solution. Too trivial? Thanks
a=b=1 isn't a solution. That would mean n=89, but 89 is not a multiple of 40 or 49
9:10 I don't think this argument is correct. Just because every 40x49 rectangle gives a zero sum doesn't mean the whole nxn square should.
By analogy, consider a row of n boxes labeled 1 through n, with box i assigned e^(i*pi/5). Then any five consecutive boxes sum to 0 but the sum in all boxes is only 0 if 5 divides n.
Every 40x40 and 49x49 square gives you sum 0. Because the nxn square is partitioned into 40x40 and 49x49 squares, you have that the total sum is 0.
Every sum of in the squares of size 40x40 and 49x49 sum to 0
Given that problem stated the board is partitioned in squares of these sizes[and the sum is commutative and associative]:
sum of all tiles = sum of (sum of the tiles in a square... which we know is 0) for each square;
On your analogy, any five consecutive boxes sum to 0. If we add an assumption on your analogy that the row of n boxes can be partitioned in groups of five consecutive boxes(just like the problem said the NxN square can be partitioned into squares of 40x40 and 49x49). Then we know the sum of all boxes is 0.
A partition of a set X is a set Y of disjoint subsets of X which, form X under union. Partitions are useful because every element once(else the union would not form X) and only once(else there would be two different sets Y1 and Y2 which aren't disjoint), so when summing you can group by set in the partition.
sum over (set X) = sum (sum over Y) for each Y in partition of X
TL DR: your analogy would have worked if you said the row of boxes could be partitioned into groups of 5 consecutive boxes.
(curiously... they don't need to be consecutive)
A rare video where I understand nothing 😅
+ means addition
One of the most smart Math. turtor❤ in tube😂👍
I like broccoli
Your post is reported for being spam. Don't make another one like it.
prof. penn wears a broccoli@@forcelifeforce
I have a Math PhD and I am still scratching my head here. Seems like one can skip straight to 13:07 with basic counting arguments, no?
Update: See proof in separate comment. 🟩
I thought the same, why bother the detour of roots of unity?
How do you conclude that either 40|n or 49|n with a basic counting argument?
Same here but my PhD was 30 years ago and I'm a bit rusty. The construction to me makes several assumptions about the form of the solution which (to me at least) are far from obvious.
He said at the very beginning he would use one of his favorite tricks. Was it a trick to show that the square sides have to be of the form 40a+49b? 🤷♂️
I don't see how such an argument would be possible. I guess you could try to count along each column and row, but you'd have to account for not necessarily having the same number of each square along each column and row. I don't know if I could think of enough to be convinced a priori that some weird stuttered pattern couldn't break basic arguments. What makes the roots of unity argument elegant is that it reduces everything to complex arithmetic. Since complex addition and multiplication are commutative, the arrangements in which the squares are placed no longer really matters. It rules out shenanigans by putting you in a nice, friendly abelian world!